Can you solve with showing steps please Vant hoff Factor 2

C. Measuring the Freezing Point of the Solution (lonic Compounds) van\'t Hoff Factor Trial 1 Trial 2 Mass of Solvent (Same for both trials) Mass of Unknown 12.05a 12.055 0.089 -WA--. NIA- Total Mass Unknown added 0.186 Freezing Point from Cooling Curve - o.OT C AT (From Part A to Part C) Molality of Solute Molar Mass of Solute Average Molar Mass of Solute

Solution

Ans. Trial 1: Depression in freezing point of the solution is given by-

                        dT_f = i K_f m             - equation 1

            where, i = Van’t Hoff factor. [ i = 2, given]

                        K_f = molal freezing point depression constant (of water) = 1.86⁰C / m

                        m = molality of the solution

                        dT_f = Freezing point of pure solvent – Freezing point of solution

# # Depression in freezing point, dT_f =

Freezing point of pure solvent – Freezing point of solution

= 0.0⁰C – (-0.07⁰C) = 0.07⁰C

# Let the molality of solution be X molal

Putting the values in equation 1-

            0.07⁰C = 2 x (1.86 ⁰C m^-1) x (X)m

            Or, X = 0.07⁰C / (3.72⁰C)

            Hence, X = 0.01882

Therefore, molality = X m = 0.01882 m

# Molality = Moles of solute / Mass of solvent in kg

Let the number of moles of solute in solution be n

            Or, 0.01882 mol / kg = n / 0.01205 kg

            Or, n = (0.01882 mol / kg) x 0.01205 kg

            Hence, n = 0.000226781 mol

# Molar mass = Mass / Moles

Or, molar mass = 0.089 g / 0.000226781 mol = 392.449 g/ mol

Therefore, molar mass of the unknown = 392.449 g/mol

# Its assumed that all the sample if transferred to the solvent for preparing the solution.

Trial 2: # Depression in freezing point, dT_f = 0.0⁰C – (-0.47⁰C) = 0.47⁰C

# Let the molality of solution be X molal

Putting the values in equation 1-

            0.47⁰C = 2 x (1.86 ⁰C m^-1) x (X)m

            Or, X = 0.47⁰C / (3.72⁰C)

            Hence, X = 0.1263

Therefore, molality = X m = 0.1263 m

# Molality = Moles of solute / Mass of solvent in kg

Let the number of moles of solute in solution be n

            Or, 0.1263 mol / kg = n / 0.01205 kg

            Or, n = (0.1263 mol / kg) x 0.01205 kg

            Hence, n = 0.001521915 mol

# Molar mass = Mass / Moles

Or, molar mass = 0.186 g / 0.001521915 mol = 122.21 g/ mol

Therefore, molar mass of the unknown = 122.21 g/mol

# Average molar mass = (Molar mass in trial 1 + trial 2) / 2

                                    = (392.45 g/mol + 122.21 g/mol) / 2

                                    = 257.33 g/ mol

Note: The difference in molar mass of the two trials is very large. Please recheck your experimental values to get comparable values of molar mass in the two trials.